Problem: Let $g(x)=\sin(3x)$, for $0 \leq x \leq \pi$. Where does $g$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=\dfrac{\pi}{6}$ (Choice B) B $x=\dfrac{\pi}{2}$ (Choice C) C $x=\dfrac{5\pi}{6}$ (Choice D) D $g$ has no critical points.
Answer: A critical point of $g$ is a point in the domain of $g$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $g$, let's find its derivative. $\begin{aligned} g'(x)&=\dfrac{d}{dx}\left[ \sin(3x) \right] \\\\ &=\cos(3x) \cdot \dfrac{d}{dx}[3x] \\\\ &=3\cos(3x) \end{aligned}$ Now let's look for $x$ -values where $g'$ is zero or undefined. $3\cos(3x)=0$ at $x=\dfrac{\pi}{6}$, $x=\dfrac{\pi}{2}$, and $x=\dfrac{5\pi}{6}$. $3\cos(3x)$ is never undefined, so $g'$ is never undefined. In conclusion, these are the $x$ -values where $g$ has critical points: $x=\dfrac{\pi}{6}$ $x=\dfrac{\pi}{2}$ $x=\dfrac{5\pi}{6}$